Difference between revisions of "Proofs without words"
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<!-- | <!-- | ||
− | + | Links: | |
− | + | http://usamts.org/Gallery/G_Gallery.php | |
− | + | http://mathoverflow.net/questions/8846/proofs-without-words | |
− | + | http://gurmeet.net/computer-science/mathematical-recreations-proofs-without-words/ | |
− | |||
hockey stick? | hockey stick? | ||
+ | Catalan numbers / reflection principle | ||
+ | geometric series: http://mahalanobis.twoday.net/stories/3472911/ | ||
+ | Fibonacci: sum_{i=1}^{2n-1} F_iF_{i+1} = F_{2n}^2 | ||
+ | sum of squares, http://legacy.lclark.edu/~mathsci/invitation.pdf | ||
+ | weighted sum of triangle numbers: http://legacy.lclark.edu/~mathsci/weightedsum.pdf | ||
+ | pentagonal/octagonal numbers: http://legacy.lclark.edu/~mathsci/octnos.pdf | ||
geometry: | geometry: | ||
Pythagorean Theorem (multiple ways - 5?) | Pythagorean Theorem (multiple ways - 5?) | ||
− | + | smallest triangle in angle (Engel 315 5) | |
− | smallest triangle in angle | ||
− | |||
equilateral triangle in three lines | equilateral triangle in three lines | ||
− | |||
Fermat point problem | Fermat point problem | ||
− | |||
Fagnano problem | Fagnano problem | ||
− | |||
− | |||
− | |||
regular heptagon identity? | regular heptagon identity? | ||
− | |||
− | |||
− | |||
− | |||
− | |||
Law of Cosines | Law of Cosines | ||
− | |||
Law of Sines | Law of Sines | ||
− | |||
radical axes/homotheties? | radical axes/homotheties? | ||
+ | Engel 313 E12, 316 6, 11, 323 4, 326/7 | ||
+ | arbelos/salinon: http://legacy.lclark.edu/~mathsci/arbelos.pdf, http://legacy.lclark.edu/~mathsci/salinon.pdf | ||
+ | abc/4R: http://legacy.lclark.edu/~mathsci/euler.pdf | ||
+ | area of triangle xy: http://legacy.lclark.edu/~mathsci/area-rt-tri.pdf | ||
misc. | misc. | ||
3D QM-GM | 3D QM-GM | ||
− | + | AM-GM from http://jwilson.coe.uga.edu/emt725/AMGM/AMGM.5.html | |
− | |||
− | |||
Minkowski's inequality | Minkowski's inequality | ||
− | |||
rt(a^2 - ab + b^2) rt(b^2 - bc + c^2) < rt(a^2 + ac + c^2) | rt(a^2 - ab + b^2) rt(b^2 - bc + c^2) < rt(a^2 + ac + c^2) | ||
− | |||
− | |||
− | |||
− | |||
cardinality of interval with R | cardinality of interval with R | ||
− | + | domino tiling, per mathoverflow | |
+ | hexagon tiling with cube (aops logo) | ||
--> | --> | ||
== Summations == | == Summations == | ||
Line 55: | Line 44: | ||
int n = 6; pair shiftR = ((n+2),0); real r = 0.3; | int n = 6; pair shiftR = ((n+2),0); real r = 0.3; | ||
pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */ | pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */ | ||
− | void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } | + | void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } |
/* triangle */ | /* triangle */ | ||
draw((-r,0)--(-r,-n+1)^^(r,-n+1)--(r,0),linetype("4 4")); | draw((-r,0)--(-r,-n+1)^^(r,-n+1)--(r,0),linetype("4 4")); | ||
− | for(int i = 0; i < n; ++i) | + | for(int i = 0; i < n; ++i) |
draw((-i,-i)--(i,-i)); | draw((-i,-i)--(i,-i)); | ||
for(int i = 0; i < n; ++i) | for(int i = 0; i < n; ++i) | ||
Line 149: | Line 138: | ||
The sum of the first <math>n</math> positive integers is <math>{n+1 \choose 2}</math>.{{ref|1}}<br><br> | The sum of the first <math>n</math> positive integers is <math>{n+1 \choose 2}</math>.{{ref|1}}<br><br> | ||
+ | </center> | ||
+ | |||
+ | <center><asy>unitsize(15); defaultpen(linewidth(0.7)); | ||
+ | int n = 5; pair shiftR = (2*n + 4, 1); real r = 0.35; pen sm = fontsize(10); | ||
+ | pen colors(int i){ return rgb(i/n,0.4+i/(2n),1-i/n); } /* shading */ | ||
+ | void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } | ||
+ | |||
+ | /* draw plus/minus circle at A */ | ||
+ | void plus (pair A) { filldraw(CR(A,r),rgb(0.5,1,0.5)); MP("+",A,(0,0), sm); } | ||
+ | void minus (pair A) { filldraw(CR(A,r),rgb(1,0.5,0.5)); MP("-",A,(0,0), sm); } | ||
+ | |||
+ | /* triangle 1 */ | ||
+ | path p1 = (-1,-1/2) -- (1,-1/2); | ||
+ | draw((-n+1-r,-n+1-2*r)--(n-1+r,-n+1-2*r)); | ||
+ | for(int i = 0; i < n; ++i) { | ||
+ | for(int j = 0; j < 2*i+1; ++j) { | ||
+ | if (i % 2 == 0) | ||
+ | plus((j-i,-i)); | ||
+ | else | ||
+ | minus((j-i,-i)); | ||
+ | } | ||
+ | if (i % 2 != 0) | ||
+ | p1 = (-i-1/2,-i-3/2)--(-i-1/2,-i+1/2)--p1--(i+1/2,-i+1/2)--(i+1/2,-i-3/2); | ||
+ | plus((2*i - n + 1, -n + 1 - 4*r)); | ||
+ | } | ||
+ | p1 = p1 -- cycle; | ||
+ | draw(p1, rgb(0.85,0.85,0.85)); | ||
+ | |||
+ | /* triangle 2 */ | ||
+ | n = n+1; | ||
+ | path p2 = (-0.1,1/2) -- (0.1,1/2); | ||
+ | draw(shiftR+ (-n+1-r,-n+1-2*r)-- shiftR+ (n-1+r,-n+1-2*r)); | ||
+ | for(int i = 0; i < n; ++i) { | ||
+ | for(int j = 0; j < 2*i+1; ++j) { | ||
+ | if (i % 2 != 0) | ||
+ | plus(shiftR+ (j-i,-i)); | ||
+ | else | ||
+ | minus(shiftR+ (j-i,-i)); | ||
+ | } | ||
+ | if (i % 2 == 0) | ||
+ | p2 = (-i-1/2,-i-3/2)--(-i-1/2,-i+1/2)--p2--(i+1/2,-i+1/2)--(i+1/2,-i-3/2); | ||
+ | plus(shiftR+ (2*i - n + 1, -n + 1 - 4*r)); | ||
+ | } | ||
+ | p2 = p2 -- cycle; | ||
+ | draw(shift(shiftR)*p2, rgb(0.85,0.85,0.85)); | ||
+ | htick(shiftR+(-n,r),shiftR+(-n,-n+1-r)); label("$n$",shiftR+(-n,(-n+1)/2),W,fontsize(10)); | ||
+ | </asy> <br> | ||
+ | |||
+ | The alternating sum of the first <math>n</math> odd natural numbers is <math>\sum_{k=1}^n (-1)^{n-k}(2k-1) = n</math>. ([http://www.math.hmc.edu/~benjamin/papers/OddPWW.pdf Source])<br><br> | ||
</center> | </center> | ||
Line 155: | Line 193: | ||
int lvl(int i){ return ceil(((8*i+9)^.5-1)/2); } | int lvl(int i){ return ceil(((8*i+9)^.5-1)/2); } | ||
− | pen colors(int i){ return rgb(0. | + | pen colors(int i){ return rgb(0.5-lvl(i)/5,0.3+lvl(i)/7,1-lvl(i)/6); } /* shading */ |
void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } | void htick(pair A, pair B,pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } | ||
Line 184: | Line 222: | ||
</asy><br> | </asy><br> | ||
− | [[Nichomauss' Theorem]]: <math>n | + | [[Nichomauss' Theorem]]: the sum of the first <math>n</math> cubes can be written as the square of the sum of the first <math>n</math> integers, a statement that can be written as <math>1^3+2^3+...+n^3=(1+2+...+n)^2</math>. <br><br> |
Here, we use the same re-arrangement as the first proof on this page (the sum of first odd integers is a square). Here's another re-arrangement to see this: | Here, we use the same re-arrangement as the first proof on this page (the sum of first odd integers is a square). Here's another re-arrangement to see this: | ||
<br><asy>defaultpen(linewidth(0.7)); unitsize(15); pen heavy = linewidth(2); | <br><asy>defaultpen(linewidth(0.7)); unitsize(15); pen heavy = linewidth(2); | ||
Line 196: | Line 234: | ||
int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */ | int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */ | ||
− | pen colors(int i) { return rgb(0. | + | pen colors(int i) { return rgb(0.5-lvl(i)/5,0.3+lvl(i)/7,1-lvl(i)/6); } /* shading */ |
/* draw tick line with label, segment between A and B */ | /* draw tick line with label, segment between A and B */ | ||
Line 261: | Line 299: | ||
real h = 0.6; // scale factor of diagram | real h = 0.6; // scale factor of diagram | ||
pair shiftR1 = (n*h+1,0), // middle diagram shift offset | pair shiftR1 = (n*h+1,0), // middle diagram shift offset | ||
− | shiftR2 = shiftR1 + (n*h+1, | + | shiftR2 = shiftR1 + (n*h+1,-1); // right diagram shift offset |
int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */ | int lvl(int i) { return ceil(((8*i+9)^.5-1)/2); } /* return level of square i */ | ||
− | pen colors(int i) { return rgb(0. | + | pen colors(int i) { return rgb(0.5-lvl(i)/5,0.3+lvl(i)/7,1-lvl(i)/6); } /* shading */ |
/* draw tick line with label, segment between A and B */ | /* draw tick line with label, segment between A and B */ | ||
Line 350: | Line 388: | ||
} | } | ||
} | } | ||
− | </asy></center> | + | </asy><br>An animated version of this proof can be found in [http://usamts.org/Gallery/G_Gallery.php this gallery].<br><br></center> |
+ | |||
+ | |||
+ | <center><asy> | ||
+ | // To change the value of n shown, edit the line "int n = 5;" to whichever desired value of n. | ||
+ | // To edit the size of the diagram, change the line unitsize(15); to the desired size. | ||
+ | |||
+ | unitsize(15); defaultpen(linewidth(0.7)); | ||
+ | |||
+ | int n = 5; // nth pentagonal number | ||
+ | real r = 0.2; // dot radius | ||
+ | pen p0 = red, p1 = rgb(1,1,0.5), p2 = rgb(0.5,1,0.5), p3 = rgb(0.5,0.5,1); // dot color | ||
+ | |||
+ | // return the coordinate of the ith point of a regular pentagon with radius s | ||
+ | // 0 <= i <= 4, and pentagonalPt(0,s) = (0,0) | ||
+ | pair pentagonalPt(int i, real s) { | ||
+ | return s * (dir(i*72+216+18) - dir(216+18)); | ||
+ | } | ||
+ | |||
+ | // draw triangles | ||
+ | filldraw(pentagonalPt(4,1)--pentagonalPt(4,n-1)--pentagonalPt(3,n-1)+(pentagonalPt(4,n-1)-pentagonalPt(3,n-1))/(n-1)--cycle, p1, linewidth(1)); | ||
+ | filldraw(pentagonalPt(3,1)--pentagonalPt(3,n-1)--pentagonalPt(2,n-1)+(pentagonalPt(3,n-1)-pentagonalPt(2,n-1))/(n-1)--cycle, p2, linewidth(1)); | ||
+ | filldraw(pentagonalPt(2,1)--pentagonalPt(2,n-1)--pentagonalPt(1,n-1)+(pentagonalPt(2,n-1)-pentagonalPt(1,n-1))/(n-1)--cycle, p3, linewidth(1)); | ||
+ | |||
+ | draw(pentagonalPt(1, n-1)--(0,0)--pentagonalPt(4, n-1)); | ||
+ | for(int i = 1; i < n; ++i) { | ||
+ | for(int k = 2; k <= 4; ++k) { | ||
+ | draw(pentagonalPt(k, i) -- pentagonalPt(k-1, i), linetype("2 2")); | ||
+ | for(int j = 0; j <= i; ++j) { | ||
+ | filldraw(circle(pentagonalPt(k, i) + j*(pentagonalPt(k-1, i)-pentagonalPt(k, i))/i, r), p0); | ||
+ | } | ||
+ | } | ||
+ | } | ||
+ | filldraw(circle((0,0),r), p0); | ||
+ | </asy><br> | ||
+ | The <math>n</math>th [[pentagonal number]] is the sum of <math>n</math> and three times the <math>n-1</math>th [[triangular number]]. <br> If <math>P_n</math> denotes the <math>n</math>th pentagonal number, then <math>P_n = 3T_{n-1}+n</math>. | ||
+ | <br><br></center> | ||
<center><asy>defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); | <center><asy>defaultpen(linewidth(0.7)); unitsize(15); pen sm = fontsize(10); | ||
Line 382: | Line 456: | ||
<center>[[#toc|Back to Top]]</center> | <center>[[#toc|Back to Top]]</center> | ||
− | === Geometric | + | === Geometric Series === |
<center><asy>defaultpen(linewidth(0.7)); unitsize(15); | <center><asy>defaultpen(linewidth(0.7)); unitsize(15); | ||
int n = 10; /* # of iterations */ | int n = 10; /* # of iterations */ | ||
Line 460: | Line 534: | ||
</asy><br> | </asy><br> | ||
− | Another proof of the identity <math>\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \cdots = \frac 13</math>. <br><br> | + | Another proof of the identity <math>\frac 14 + \frac {1}{4^2} + \frac {1}{4^3} + \frac {1}{4^4}+\cdots = \frac 13</math>. <br><br> |
</center> | </center> | ||
Line 515: | Line 589: | ||
} | } | ||
</asy><br><br> | </asy><br><br> | ||
− | The [[arithmetic-geometric series]] <math>\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{ | + | The [[arithmetic-geometric series]] <math>\sum_{n=1}^{\infty} nr^n = \sum_{n=1}^{\infty} \sum_{i=n}^{\infty} r^i = \sum_{n=1}^{\infty} \frac{r^{n}}{1-r} = \frac{r}{(1-r)^2}</math>, also known as Gabriel's staircase.{{ref|2}}<br><br></center> |
<center>[[#toc|Back to Top]]</center> | <center>[[#toc|Back to Top]]</center> | ||
Line 613: | Line 687: | ||
Another proof of the [[Pythagorean Theorem]]; the left-hand diagram suggests the identity <math>c^2 = (a+b)^2 - 4 \cdot \frac{ab}2 = a^2 + b^2</math>, and the right-hand diagram offers another re-arrangement proof.<br><br></center> | Another proof of the [[Pythagorean Theorem]]; the left-hand diagram suggests the identity <math>c^2 = (a+b)^2 - 4 \cdot \frac{ab}2 = a^2 + b^2</math>, and the right-hand diagram offers another re-arrangement proof.<br><br></center> | ||
+ | |||
+ | <center><asy> | ||
+ | import graph; defaultpen(linewidth(0.7)); unitsize(15); pen ds=black, ls = linetype("2 2"); | ||
+ | pen fueaev=rgb(1,0.4,0.4), ffffea=rgb(1,1,0.4), evfuea=rgb(0.4,1,0.4), fffuev=rgb(1,0.5,0.3), fzfzfz=rgb(0.8,0.8,0.8), evevfz=rgb(0.4,0.4,1); | ||
+ | |||
+ | fill((0,0)--(1.5,2)--(-0.5,3.5)--(-2,1.5)--cycle,fueaev); fill((1.5,2)--(4.17,0)--(6.17,2.67)--(3.5,4.67)--cycle,fueaev); fill((4.17,0)--(0,0)--(0,-4.17)--(4.17,-4.17)--cycle,fueaev); fill((1.5,2)--(2,2.67)--(2,1.63)--cycle,ffffea); fill((2,2.67)--(2,1.63)--(4.17,0)--(6.17,2.67)--cycle,evfuea); fill((0,3.13)--(0,0)--(1.5,2)--cycle,fffuev); fill((-0.5,3.5)--(0,3.13)--(0,0)--(-2,1.5)--cycle,fueaev); fill((0,0)--(1.5,2)--(4.17,0)--cycle,fzfzfz); fill((2,2.67)--(3.5,4.67)--(6.17,2.67)--cycle,evevfz); fill((0,-4.17)--(1.5,-2.17)--(0,-1.04)--cycle,fffuev); fill((1.5,-2.17)--(0,-4.17)--(4.17,-4.17)--cycle,evevfz); fill((4.17,0)--(3.67,-0.67)--(4.17,-1.04)--cycle,white); fill((3.67,-0.67)--(2.17,-2.67)--(4.17,-4.17)--(4.17,-1.04)--cycle,fueaev); fill((0,0)--(0,-1.04)--(2.17,-2.67)--(4.17,0)--cycle,evfuea); fill((4.17,0)--(3.67,-0.67)--(4.17,-1.04)--cycle, ffffea); | ||
+ | |||
+ | draw((0,0)--(1.5,2)); draw((0,0)--(4.17,0)); draw((1.5,2)--(4.17,0)); draw((0,0)--(1.5,2)); draw((1.5,2)--(-0.5,3.5)); draw((-0.5,3.5)--(-2,1.5)); draw((-2,1.5)--(0,0)); draw((1.5,2)--(4.17,0)); draw((4.17,0)--(6.17,2.67)); draw((6.17,2.67)--(3.5,4.67)); draw((3.5,4.67)--(1.5,2)); draw((4.17,0)--(0,0)); draw((0,0)--(0,-4.17)); draw((0,-4.17)--(4.17,-4.17)); draw((4.17,-4.17)--(4.17,0)); draw((0,0)--(-2,1.5)); draw((-2,1.5)--(-0.5,3.5)); draw((-0.5,3.5)--(1.5,2)); draw((0,0)--(0,-4.17)); draw((0,-4.17)--(4.17,-4.17)); draw((4.17,-4.17)--(4.17,0)); draw((4.17,0)--(6.17,2.67)); draw((6.17,2.67)--(3.5,4.67)); draw((3.5,4.67)--(1.5,2)); draw((0,-4.17)--(0,0)); draw((0,0)--(0,3.13)); draw((1.5,2)--(2,2.67)); draw((2,2.67)--(2,1.63)); draw((2,1.63)--(1.5,2)); draw((2,2.67)--(2,1.63)); draw((2,1.63)--(4.17,0)); draw((4.17,0)--(6.17,2.67)); draw((6.17,2.67)--(2,2.67)); draw((0,3.13)--(0,0)); draw((0,0)--(1.5,2)); draw((1.5,2)--(0,3.13)); draw((-0.5,3.5)--(0,3.13)); draw((0,3.13)--(0,0)); draw((0,0)--(-2,1.5)); draw((-2,1.5)--(-0.5,3.5)); draw((0,0)--(1.5,2)); draw((1.5,2)--(4.17,0)); draw((4.17,0)--(0,0)); draw((2,2.67)--(3.5,4.67)); draw((3.5,4.67)--(6.17,2.67)); draw((6.17,2.67)--(2,2.67)); draw((0,-1.04)--(4.17,-4.17)); draw((0,-4.17)--(1.5,-2.17)); draw((1.5,-2.17)--(0,-1.04)); draw((0,-1.04)--(0,-4.17)); draw((1.5,-2.17)--(0,-4.17)); draw((0,-4.17)--(4.17,-4.17)); draw((4.17,-4.17)--(1.5,-2.17)); draw((4.17,0)--(3.67,-0.67)); draw((3.67,-0.67)--(4.17,-1.04)); draw((4.17,-1.04)--(4.17,0)); draw((3.67,-0.67)--(2.17,-2.67)); draw((2.17,-2.67)--(4.17,-4.17)); draw((4.17,-4.17)--(4.17,-1.04)); draw((4.17,-1.04)--(3.67,-0.67)); draw((0,0)--(0,-1.04)); draw((0,-1.04)--(2.17,-2.67)); draw((2.17,-2.67)--(4.17,0)); draw((4.17,0)--(0,0)); | ||
+ | |||
+ | draw((-0.5,3.5)--(1.5,6.17),ls); | ||
+ | draw((1.5,6.17)--(3.5,4.67),ls); | ||
+ | draw((0,3.13)--(0,4.17),ls); | ||
+ | draw((1.5,6.17)--(1.5,2),ls); | ||
+ | draw((2,5.79)--(2,2.67),ls); | ||
+ | draw(rightanglemark((0,0),(1.5,2),(4.17,0))); | ||
+ | </asy><br> | ||
+ | |||
+ | A dissection proof of the [[Pythagorean Theorem]].{{ref|6}} ([http://www.cut-the-knot.org/pythagoras/PythLoomis25G.shtml Cut-the-knot])<br><br></center> | ||
<center><asy> | <center><asy> | ||
Line 627: | Line 719: | ||
</asy><br> | </asy><br> | ||
− | COMING: The last | + | COMING: The last proof of the [[Pythagorean Theorem]] we shall present on this page, this one by dissection.<br><br></center> |
+ | |||
+ | <center><asy> | ||
+ | defaultpen(linewidth(0.7)+fontsize(10)); unitsize(15); real a = 3.6, b = 4.8, c = (a^2 + b^2)^.5; pair shiftR = (a+b+2,0); pen sm = fontsize(10), heavy = linewidth(1); | ||
+ | void htick(pair A, pair B, pair ticklength = (0.15,0)){ draw(A--B ^^ A-ticklength--A+ticklength ^^ B-ticklength--B+ticklength); } | ||
+ | void makeshiftarrow(pair A, real dir, real arrowlength = 0.5){ /* Arrow option resizes */ | ||
+ | fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); | ||
+ | } | ||
+ | |||
+ | real s1 = 5, s2 = 7, s3 = 8; // triangle side lengths | ||
+ | pen c1 = rgb(0.5,0.5,1), c2 = rgb(0.5,1,0.8), c3 = rgb(0.5,1,0.5); // color pens | ||
+ | pair shiftR = (s1+1,0); // distance between two diagrams | ||
+ | |||
+ | pair A=(0,0), B = (s1,0), C = intersectionpoints(Circle(A, s3), Circle(B, s2))[0], I = incenter(A,B,C), D = foot(I,B,C), E = foot(I,A,C), F = foot(I,A,B); | ||
+ | |||
+ | // draw left diagram | ||
+ | filldraw(A--I--B--cycle, c1, heavy); filldraw(B--I--C--cycle, c2, heavy); filldraw(A--I--C--cycle, c3, heavy); | ||
+ | dot(I); draw(incircle(A,B,C), heavy); draw(I--D, linetype("2 2")); draw(I--E, linetype("2 2")); draw(I--F, linetype("2 2")); | ||
+ | label("$a$",(A+B)/2,S); label("$b$",(B+C)/2,NE); label("$c$",(A+C)/2,NW); label("$r$",(I+F)/2,W); | ||
+ | draw(rightanglemark(I,F,A)); draw(rightanglemark(I,D,B)); draw(rightanglemark(I,E,C)); | ||
+ | |||
+ | pair reflectC = C + 2*(foot(C,A,I) - C), reflectI = (reflectC.x - I.x, I.y); | ||
+ | |||
+ | // draw right diagram | ||
+ | filldraw(shift(shiftR)*(A--I--B--cycle), c1, heavy); | ||
+ | filldraw(shift(shiftR+B)*(A--reflectI--reflectC--cycle), c3, heavy); | ||
+ | filldraw(shift(shiftR)*(I--B--B+reflectI--cycle), c2, heavy); | ||
+ | // dashed perpendiculars; arbitrary coding here | ||
+ | draw(shift(shiftR)*(I--F), linetype("2 2")); | ||
+ | draw(shift(shiftR)*(B--B+(0,inradius(A,B,C))), linetype("2 2")); | ||
+ | draw(shift(shiftR+B)*(reflectI--foot(reflectI,A,B)), linetype("2 2")); | ||
+ | draw(shift(shiftR)*rightanglemark(I,F,A)); draw(shift(shiftR)*rightanglemark(B,B+(0,inradius(A,B,C)),I)); draw(shift(shiftR+B)*rightanglemark(reflectI,foot(reflectI,A,B),A)); | ||
+ | label("$a$",shiftR+(A+B)/2,S); label("$b$",shiftR+(I+B+reflectI)/2,N); label("$c$",shiftR+B+(A+reflectC)/2,S); label("$r$",shiftR+(I+F)/2,W); | ||
+ | </asy> | ||
+ | |||
+ | The area of a triangle is given by <math>A = \frac{1}{2} \cdot r \cdot (a+b+c) = rs</math>, where <math>r</math> is the [[inradius]] and <math>s</math> is the [[semiperimeter]].{{ref|10}} <br>(Comment: we do not need to re-arrange the triangles to a trapezoid to see this, but this re-arrangement works due to alternate interior angles/angle bisector properties of the incenter.)<br><br></center> | ||
+ | |||
+ | <center><asy>unitsize(15); defaultpen(linewidth(0.7) + fontsize(10)); pen heavy = linewidth(1); | ||
+ | real a = 4.5, b = 2.5, c = 2, d = 4; | ||
+ | pair A = (a,b), B = (a+c,b+d), C = (c,d), D = IP(B--B+2*(A-B), (0,0)--(a,0)), F = IP(B--B+2*(C-B), (0,0)--(0,d)), G = IP(B--D,(0,d)--(a+c,d)), H = IP(B--F,(a,0)--(a,b+d)), shiftR = (a+c+1,0), shiftR2 = 2*shiftR; | ||
+ | |||
+ | // left diagram | ||
+ | filldraw((0,0)--(a,0)--(a,d)--(0,d)--cycle, rgb(0.5, 1, 0.5)); draw((0,0)--A--B--C--cycle); draw(shift((a,d))*xscale(c)*yscale(b)*unitsquare); draw(A--D ^^ C--F, linetype("2 2")); | ||
+ | draw((0,0)--(a,0)--(a,b)--cycle, heavy); draw(shift(C)*((0,0)--(a,0)--(a,b)--cycle), heavy); | ||
+ | label("$a$",(a/2,0),S); label("$d$",(0,d/2),W); label("$b$",B-(0,b/2),E); label("$c$",B-(c/2,0),N); label("$(a,b)$",(a,b),SE,fontsize(8)); label("$(c,d)$",(c,d),NW,fontsize(8)); | ||
+ | |||
+ | // middle diagram | ||
+ | filldraw(shift(shiftR)*((0,0)--A--(a,d)--G--B--C--(0,d)--cycle), rgb(0.5, 1, 0.5)); filldraw(shift(shiftR)*(G--(a+c,d)--B--cycle), rgb(1,0.5,0.5)); draw(shift(shiftR)*((0,0)--(a,0)--(a,d)--(0,d)--cycle)); draw(shift(shiftR+(a,d))*xscale(c)*yscale(b)*unitsquare); draw(shift(shiftR)*(A--D ^^ C--F), linetype("2 2")); | ||
+ | draw(shift(shiftR)*((0,0)--(0,d)--(c,d)--cycle), heavy); draw(shift(shiftR+A)*((0,0)--(0,d)--(c,d)--cycle), heavy); | ||
+ | label("$a$",shiftR+(a/2,0),S); label("$d$",shiftR+(0,d/2),W); label("$b$",shiftR+B-(0,b/2),E); label("$c$",shiftR+B-(c/2,0),N); | ||
+ | |||
+ | // right diagram | ||
+ | filldraw(shift(shiftR2)*((0,0)--A--G--(a,d)--H--C--cycle), rgb(0.5, 1, 0.5)); | ||
+ | filldraw(shift(shiftR2)*(G--(a,d)--H--B--cycle), rgb(0.3, 0.9, 0.3)); filldraw(shift(shiftR2)*(G--(a+c,d)--B--cycle), rgb(1,0.5,0.5)); filldraw(shift(shiftR2)*(H--(a,b+d)--B--cycle), rgb(1,0.5,0.5)); draw(shift(shiftR2)*((0,0)--(a,0)--(a,d)--(0,d)--cycle)); draw(shift(shiftR2)*(A--D ^^ C--F), linetype("2 2")); | ||
+ | label("$a$",shiftR2+(a/2,0),S); label("$d$",shiftR2+(0,d/2),W); label("$b$",shiftR2+B-(0,b/2),E); label("$c$",shiftR2+B-(c/2,0),N); label("$\overrightarrow{(a,b)}$",shiftR2+(a,b),SE,fontsize(8)); label("$\overrightarrow{(c,d)}$",shiftR2+(c,d),NW,fontsize(8)); | ||
+ | </asy> | ||
+ | The area of a parallelogram with adjacent side vectors <math>\overrightarrow{(a,b)}, \overrightarrow{(c,d)}</math> is given by <math>\overrightarrow{(a,b)} \times \overrightarrow{(c,d)} = ad-bc</math>.<br><br></center> | ||
+ | |||
+ | <center><asy> | ||
+ | defaultpen(linewidth(0.7)); unitsize(15); | ||
+ | |||
+ | real r = 3.5; // radius | ||
+ | pair shiftL = (-2.5*r,0); // distance between 2 diagrams | ||
+ | |||
+ | /* returns the vertex of the interior equilateral triangle with one edge shared with the dodecagon */ | ||
+ | pair dodecagonPt(int i) { | ||
+ | return r*dir(i*360/12) + rotate(60)*(r*(dir((i+1)*360/12) - dir(i*360/12))); | ||
+ | } | ||
+ | |||
+ | /* left diagram */ | ||
+ | path dodecagon = shiftL+(r,0)--shiftL+r*dir(30); | ||
+ | for(int i = 1; i < 12; ++i) | ||
+ | dodecagon = dodecagon--shiftL+r*dir(i*30); | ||
+ | dodecagon = dodecagon--cycle; | ||
+ | filldraw(dodecagon, rgb(0.5,1,0.5)); | ||
+ | draw(Circle(shiftL, r), linetype("2 2")); | ||
+ | dot((0,0)); | ||
+ | draw(shiftL--shiftL+(r,0)); | ||
+ | label("$R$",shiftL+(r/2,0),S); | ||
+ | |||
+ | /* right diagram */ | ||
+ | for(int i = 0; i < 9; ++i) { | ||
+ | filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); | ||
+ | filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0,0.8,0)); | ||
+ | filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.8,0.8,0)); | ||
+ | if (i % 3 == 1) { | ||
+ | filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir((i+1)*360/12)--cycle, rgb(0.8,0.8,0)); | ||
+ | filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir(i*360/12)--r*dir(floor(i/3)*90)--cycle, rgb(0,0.8,0)); | ||
+ | filldraw(r*2^.5*dir(floor(i/3)*90+45)--r*dir((i+1)*360/12)--r*dir(floor(i/3)*90+90)--cycle, rgb(0,0.8,0)); | ||
+ | } | ||
+ | } | ||
+ | for(int i = 9; i < 12; ++i) { | ||
+ | filldraw((0,0)--r*dir(i*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype("2 2")); | ||
+ | filldraw((0,0)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(0.5,1,0.5), linetype("2 2")); | ||
+ | filldraw(r*dir(i*360/12)--r*dir((i+1)*360/12)--dodecagonPt(i)--cycle, rgb(1,1,0.5), linetype("2 2")); | ||
+ | } | ||
+ | </asy><br> | ||
+ | |||
+ | The area of a [[dodecagon]] is <math>3R^2</math>, where <math>R</math> is the [[circumradius]].<br><br></center> | ||
+ | |||
<center><asy> | <center><asy> | ||
Line 656: | Line 847: | ||
In trapezoid <math>ABCD</math> with <math>\overline{AB} \parallel \overline{CD}</math>, then <math>\overline{AC} \perp \overline{BD} \Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2</math>.<br><br></center> | In trapezoid <math>ABCD</math> with <math>\overline{AB} \parallel \overline{CD}</math>, then <math>\overline{AC} \perp \overline{BD} \Longleftrightarrow AC^2 + BD^2 = (AB + CD)^2</math>.<br><br></center> | ||
+ | <center><asy> | ||
+ | // feel free to change these four points! | ||
+ | pair A = (0,0), B = (3, -2), C = (5,1), D = (1,3); | ||
+ | |||
+ | // | ||
+ | // Rest of code | ||
+ | // | ||
+ | size(200); defaultpen(linewidth(0.9)); pen lightgreen = rgb(0.6,1,0.6), lightred = rgb(1,0.6,0.6), smdash = linewidth(0.7)+linetype("2 2"); | ||
+ | |||
+ | pair E = IP(A--C,B--D), AB = (A+B)/2, BC = (B+C)/2, CD = (C+D)/2, DA = (D+A)/2, ABE = 2*AB-E, BCE = 2*BC-E, CDE = 2*CD-E, DAE = 2*DA-E; | ||
+ | path midpts = AB--BC--CD--DA--cycle; | ||
+ | filldraw(shift(-E)*scale(2)*midpts,lightgreen); filldraw(midpts,lightred); | ||
+ | draw(A--B--C--D--cycle); draw(A--C); draw(B--D); draw((A+ABE)/2--(C+BCE)/2,smdash); draw((B+ABE)/2--(D+DAE)/2,smdash); draw((B+BCE)/2--(D+CDE)/2,smdash); draw((A+DAE)/2--(C+CDE)/2,smdash); | ||
+ | |||
+ | dot(A); dot(B); dot(C); dot(D); label("$A$",A,W);label("$B$",B,S);label("$C$",C,E);label("$D$",D,N); | ||
+ | </asy><br> | ||
+ | |||
+ | [[Varignon's theorem]]: the area of the outer parallelogram is twice the area of the quadrilateral and four times the area of the midpoint parallelogram, so the midpoint parallelogram of a (convex) quadrilateral has area <math>1/2</math> of the quadrilateral. | ||
+ | </center> | ||
+ | |||
+ | Proof for Volume of a Cone: http://www.mathematische-basteleien.de/wuerfel16.gif | ||
<center>[[#toc|Back to Top]]</center> | <center>[[#toc|Back to Top]]</center> | ||
− | [[ | + | == Miscellaneous == |
+ | <center><asy> | ||
+ | import graph; size(220); defaultpen(linewidth(0.7)); Label k; k.p=fontsize(10); | ||
+ | |||
+ | real xmax = pi/2+0.5, xmin = -0.5, ymax = 1.39, ymin = -0.39, lblpt = pi/4 + 0.08; | ||
+ | |||
+ | /* f(x) = sin^2(x) */ | ||
+ | real f(real x) { return sin(x) * sin(x); } | ||
+ | string pilabel(real x) { if(x > 1) return "$\pi/2$"; else if(x > 0) return "$\pi/4$"; else return "";} | ||
+ | |||
+ | xaxis(xmin,xmax,Ticks(k, pilabel, pi/4),Arrows(6)); yaxis(ymin,ymax,Ticks(k, NoZero),Arrows(6)); | ||
+ | filldraw(graph(f,0,pi/2)--(pi/2,0)--(0,0)--cycle,gray(0.7),linewidth(1)); | ||
+ | draw((lblpt,f(lblpt))--(lblpt,1),Arrows(6)); draw((lblpt,f(lblpt))--(lblpt,0),Arrows(6)); | ||
+ | label("$\cos^2(x)$",(lblpt,f(lblpt)/2+1/2),W,fontsize(10)); | ||
+ | label("$\sin^2(x)$",(lblpt,f(lblpt)/2),E,fontsize(10)); | ||
+ | draw((0,1)--(pi/2,1),linewidth(1)); | ||
+ | </asy> | ||
+ | <math>\int_0^{\pi/2} \sin^2 x \, dx = \int_0^{\pi/2} \cos^2 x \, dx = \frac {\pi}{4}</math> from <math> | ||
+ | <br><br></center> | ||
+ | |||
+ | <center><asy> | ||
+ | import graph; size(170); defaultpen(linewidth(0.7)); Label k; k.p=fontsize(8); | ||
+ | |||
+ | real xmax = 1.35, xmin = -0.35, ymax = 1.35, ymin = -0.35, lblpt = 1/2 + 0.15, alpha = 2, epsilon = 0.015; | ||
+ | |||
+ | real f(real x) { return x^alpha; } | ||
+ | |||
+ | xaxis(xmin,xmax,Ticks(k, 0.25, NoZero),Arrows(6)); yaxis(ymin,ymax,Ticks(k, 0.25, NoZero),Arrows(6)); | ||
+ | filldraw(graph(f,0,1)--(1,0)--(0,0)--cycle,gray(0.7),linewidth(1)); | ||
+ | draw(graph(f,1,1.1),linewidth(0.7),EndArrow(4)); | ||
+ | draw((lblpt-epsilon,f(lblpt))--(0+epsilon,f(lblpt)),Arrows(4)); draw((lblpt,f(lblpt)-epsilon)--(lblpt,0+epsilon),Arrows(4)); | ||
+ | label("$x = y^{1/\alpha}$",(lblpt/2,f(lblpt)),N,fontsize(10)); | ||
+ | label("$y = x^{\alpha}$",(lblpt,f(lblpt)/2),E,fontsize(10)); | ||
+ | draw((0,1)--(1,1),linewidth(1)); | ||
+ | </asy> | ||
+ | <math>\alpha > 0 \Longrightarrow \int_0^{1} \left(x^\alpha + x^{1/\alpha}\right) \, dx = 1.</math> ([http://www.math.ubc.ca/~morey/talk/proofwowords.html Source]) | ||
+ | <br><br></center> | ||
− | |||
<center><asy>unitsize(15); defaultpen(linewidth(0.7)); | <center><asy>unitsize(15); defaultpen(linewidth(0.7)); | ||
real a=2.5,b=5,s=a+b; pen colors[] = {rgb(0.9,0.2,0.2), rgb(0.2,0.9,0.2), rgb(0.2,0.2,0.9)}; pen sm = fontsize(8); | real a=2.5,b=5,s=a+b; pen colors[] = {rgb(0.9,0.2,0.2), rgb(0.2,0.9,0.2), rgb(0.2,0.2,0.9)}; pen sm = fontsize(8); | ||
Line 713: | Line 960: | ||
The [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality]].{{ref|5}}<br><br> | The [[Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality]].{{ref|5}}<br><br> | ||
</center> | </center> | ||
+ | |||
<center><asy> unitsize(15); defaultpen(linewidth(0.7)); | <center><asy> unitsize(15); defaultpen(linewidth(0.7)); | ||
Line 746: | Line 994: | ||
[[Fermat's Little Theorem]]: <math>a^p \equiv a \pmod{p}</math> for <math>\text{gcd}\,(a,p) = 1</math> (above <math>a=2,p=3</math>).<br><br> | [[Fermat's Little Theorem]]: <math>a^p \equiv a \pmod{p}</math> for <math>\text{gcd}\,(a,p) = 1</math> (above <math>a=2,p=3</math>).<br><br> | ||
</center> | </center> | ||
+ | <center><asy> | ||
+ | defaultpen(linewidth(0.7)); unitsize(30); | ||
+ | |||
+ | real r = 0.2; | ||
+ | pair endPt1 = (-2.5,-1), endPt2 = (2.5,-1); // endpoints of R^1 line | ||
+ | real projections[] = {-1.5, -0.75, 1, 2.2}; // x-coordinates of steoreographic projections on y=-1 | ||
+ | |||
+ | void makeshiftarrow(pair A, real dir, real arrowlength = r){ /* Arrow option resizes, so draw makeshift arrows */ | ||
+ | fill(A--A+arrowlength*expi(dir+pi/8)--A+arrowlength*expi(dir-pi/8)--cycle); | ||
+ | } | ||
+ | |||
+ | draw(endPt1 -- endPt2); | ||
+ | draw(unitcircle); | ||
+ | for(int i = 0; i < projections.length; ++i) { | ||
+ | draw((0,1) -- (projections[i],-1), linetype("2 2")); | ||
+ | dot((projections[i],-1), Fill(red)); | ||
+ | dot(OP((0,1) -- (projections[i],-1),unitcircle), Fill(yellow)); | ||
+ | } | ||
+ | dot((0,1), Fill(green)); | ||
+ | |||
+ | makeshiftarrow(endPt1, 0); makeshiftarrow(endPt2, pi); | ||
+ | </asy><asy> | ||
+ | import three; defaultpen(linewidth(0.7)); unitsize(20); | ||
+ | |||
+ | currentprojection = orthographic(0.3,-2,0.6); | ||
+ | |||
+ | pen gridpen = dotted; | ||
+ | int gridmin = -2, gridmax = 2, gridlines = 6; | ||
+ | pair projections[] = {(1,-1),(1,-2),(2,-2),(2,-1)}; | ||
+ | triple IPs[] = new triple[4]; | ||
+ | draw((1,-1,-1)--(1,-2,-1)--(2,-2,-1)--(2,-1,-1)--cycle); | ||
+ | |||
+ | for(int i = 0; i < gridlines; ++i) { | ||
+ | draw((gridmin,gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),-1)--(gridmax,gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),-1), gridpen); | ||
+ | draw((gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),gridmin,-1)--(gridmin+(gridmax-gridmin)*(i+1)/(gridlines+1),gridmax,-1), gridpen); | ||
+ | } | ||
+ | |||
+ | draw(unitsphere,white); | ||
+ | draw(arc((0,0,0), (1,0,0), (-1,0,0), (0,0,1)), linetype("2 2")); | ||
+ | draw(arc((0,0,0), (-1,0,0), (1,0,0), (0,0,1))); | ||
+ | // draw(circle((0,0,0), 1, (0,1,0))); | ||
+ | |||
+ | // draw projection points | ||
+ | for(int i = 0; i < projections.length; ++i) { | ||
+ | real px = projections[i].x/2, py = projections[i].y/2, pxy = 1 + px*px + py*py; | ||
+ | draw((0,0,1)--(2*px, 2*py, -1), linetype("2 2")); | ||
+ | triple IP = (2*px/pxy, 2*py/pxy, (pxy-2)/pxy), OP = (2*px, 2*py, -1); | ||
+ | dot(IP, yellow); | ||
+ | draw(circle(IP, 0.06, (0,1,0))); | ||
+ | IPs[i] = IP; | ||
+ | dot(OP, red); | ||
+ | draw(circle(OP, 0.06, (0,1,0))); | ||
+ | if(i != 0) | ||
+ | draw(IPs[i] -- IPs[i-1], dotted); | ||
+ | } | ||
+ | draw(IPs[0]--IPs[3], dotted); | ||
+ | |||
+ | dot((0,0,1),green); | ||
+ | draw(circle((0,0,1), 0.06, (0,1,0))); | ||
+ | </asy> | ||
+ | There exists a [[homeomorphism]], the [[stereographic projection]], between the punctured hypersphere <math>S^n \setminus \{(1,\underbrace{0, \ldots, 0}_{n-1\text{ zeroes}})\}</math> and <math>\mathbb{R}^n</math> for <math>n = 1,2</math>. | ||
+ | <br><br></center> | ||
+ | |||
+ | Sum of arctangents formula: | ||
+ | |||
+ | <asy> | ||
+ | /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki, go to User:Azjps/geogebra */ | ||
+ | import graph; usepackage("amsmath"); size(13cm); | ||
+ | real labelscalefactor = 0.5; /* changes label-to-point distance */ | ||
+ | pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ | ||
+ | pen dotstyle = black; /* point style */ | ||
+ | real xmin = -0.4717093412177357, xmax = 7.405441345585962, ymin = -1.1854534297865673, ymax = 7.342957746870971; /* image dimensions */ | ||
+ | pen uququq = rgb(0.25098039215686274,0.25098039215686274,0.25098039215686274); pen aqaqaq = rgb(0.6274509803921569,0.6274509803921569,0.6274509803921569); pen qqwuqq = rgb(0.,0.39215686274509803,0.); pen cqcqcq = rgb(0.7529411764705882,0.7529411764705882,0.7529411764705882); | ||
+ | |||
+ | draw((0.,0.)--(0.,1.)--(1.,1.)--cycle); | ||
+ | draw((1.,1.)--(0.,3.)--(0.,1.)--cycle, uququq); | ||
+ | draw((0.,3.)--(6.,6.)--(1.,1.)--cycle, aqaqaq); | ||
+ | draw(arc((1.,1.),0.3101240427875472,180.,225.)--(1.,1.)--cycle, qqwuqq); | ||
+ | draw(arc((1.,1.),0.3101240427875472,116.56505117707799,180.)--(1.,1.)--cycle, qqwuqq); | ||
+ | draw(arc((1.,1.),0.3101240427875472,45.,116.56505117707799)--(1.,1.)--cycle, qqwuqq); | ||
+ | /* draw grid of horizontal/vertical lines */ | ||
+ | pen gridstyle = linewidth(0.7) + cqcqcq; real gridx = 1., gridy = 1.; /* grid intervals */ | ||
+ | for(real i = ceil(xmin/gridx)*gridx; i <= floor(xmax/gridx)*gridx; i += gridx) | ||
+ | draw((i,ymin)--(i,ymax), gridstyle); | ||
+ | for(real i = ceil(ymin/gridy)*gridy; i <= floor(ymax/gridy)*gridy; i += gridy) | ||
+ | draw((xmin,i)--(xmax,i), gridstyle); | ||
+ | /* end grid */ | ||
+ | |||
+ | /* draw figures */ | ||
+ | draw((0.,0.)--(0.,1.)); | ||
+ | draw((0.,1.)--(1.,1.)); | ||
+ | draw((1.,1.)--(0.,0.)); | ||
+ | draw((1.,1.)--(0.,3.), uququq); | ||
+ | draw((0.,3.)--(0.,1.), uququq); | ||
+ | draw((0.,1.)--(1.,1.), uququq); | ||
+ | draw((0.,3.)--(6.,6.), aqaqaq); | ||
+ | draw((6.,6.)--(1.,1.), aqaqaq); | ||
+ | draw((1.,1.)--(0.,3.), aqaqaq); | ||
+ | label("$\arctan 1 + \arctan 2 + \arctan 3 = \pi$",(1.544096936901321,0.5925910821953678),SE*labelscalefactor,fontsize(10)); | ||
+ | /* dots and labels */ | ||
+ | dot((0.,0.),linewidth(3.pt) + dotstyle); | ||
+ | dot((0.,1.),dotstyle); | ||
+ | dot((1.,1.),dotstyle); | ||
+ | dot((0.,3.),dotstyle); | ||
+ | dot((6.,6.),dotstyle); | ||
+ | clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); | ||
+ | /* end of picture */ | ||
+ | </asy> | ||
<center>[[#toc|Back to Top]]</center> | <center>[[#toc|Back to Top]]</center> |
Latest revision as of 15:07, 16 September 2024
The following demonstrate proofs of various identities and theorems using pictures, inspired from this gallery.
Summations
The sum of the first odd natural numbers is .
The sum of the first positive integers is .
The sum of the first positive integers is .[1]
The alternating sum of the first odd natural numbers is . (Source)
Nichomauss' Theorem: the sum of the first cubes can be written as the square of the sum of the first integers, a statement that can be written as .
Here, we use the same re-arrangement as the first proof on this page (the sum of first odd integers is a square). Here's another re-arrangement to see this:
This also suggests the following alternative proof:
An animated version of this proof can be found in this gallery.
The th pentagonal number is the sum of and three times the th triangular number.
If denotes the th pentagonal number, then .
The identity , where is the th Fibonacci number.
Geometric Series
The infinite geometric series .
The infinite geometric series .
The infinite geometric series .
Another proof of the identity .
The infinite geometric series .
The arithmetic-geometric series , also known as Gabriel's staircase.[2]
Geometry
The Pythagorean Theorem (first of many proofs): the left diagram shows that , and the right diagram shows a second proof by re-arranging the first diagram (the area of the shaded part is equal to , but it is also the re-arranged version of the oblique square, which has area ).[3]
Another proof of the Pythagorean Theorem (animated version).
Another proof of the Pythagorean Theorem; the left-hand diagram suggests the identity , and the right-hand diagram offers another re-arrangement proof.
A dissection proof of the Pythagorean Theorem.[6] (Cut-the-knot)
COMING: The last proof of the Pythagorean Theorem we shall present on this page, this one by dissection.
(Comment: we do not need to re-arrange the triangles to a trapezoid to see this, but this re-arrangement works due to alternate interior angles/angle bisector properties of the incenter.)
The area of a dodecagon is , where is the circumradius.
The smallest distance necessary to travel between , the x-axis, and then for is given by .[4]
In trapezoid with , then .
Varignon's theorem: the area of the outer parallelogram is twice the area of the quadrilateral and four times the area of the midpoint parallelogram, so the midpoint parallelogram of a (convex) quadrilateral has area of the quadrilateral.
Proof for Volume of a Cone: http://www.mathematische-basteleien.de/wuerfel16.gif
Miscellaneous
from . (Source)
(Source)
The Root-Mean Square-Arithmetic Mean-Geometric Mean inequality, .
The Root-Mean Square-Arithmetic Mean-Geometric Mean-Harmonic mean Inequality.[5]
Fermat's Little Theorem: for (above ).
There exists a homeomorphism, the stereographic projection, between the punctured hypersphere and for .
Sum of arctangents formula:
References
- ^ MathOverflow
- ^ Wolfram MathWorld
- ^ Attributed to the Chinese text Zhou Bi Suan Jing.
- ^ This is more of a proof without words of the AM-GM inequality ; though the lengths of the segments labeled RMS and HM can easily be verified to have values of , respectively, it might not be obvious from the diagram. It still serves as a useful graphical demonstration of the inequality.